Gravitation+SHM

I enjoyed writing down the full equation of motion for the pendulum and then after approximating it for small angles. Let $\theta$ be the angle formed by the pendulum with the vertical and $\psi$ be the angle formed by the vertical and the line joining the centre of the planet with the pendulum. Refer to the ugly figure I drew.

The modulus of the gravitational force acting on the pendulum is: $F = \frac{2GMm}{(2R + x)^2}$ Applying Pythagoras' theorem to the lower right triangle, we can establish that: $(2R + x)^2 = 4R^2\sin^2\theta + 4R^2(2 - \cos\theta)^2$ and thus $F = \frac{GMm}{2R^2(5 - 4\cos\theta)}$ Only the tangential component contributes to the motion, so we need to establish $\sin(\theta+\psi)$ . Using a bit of trigonometry, we have $\tan\psi = \frac{\sin\theta}{2 - \cos\theta}$ and so we can write the equation of motion as: $\ddot\theta + \frac{GM\sin[\theta + \arctan(\frac{\sin\theta}{2 - \cos\theta})]}{4R^3(5 - 4\cos\theta)} = 0$ For $\theta \ll 1$ , we can expand the second term and obtain: $\ddot\theta + \frac{GM}{2R^3}\theta = 0$ The factor multiplying $\theta$ is the squared angular velocity $\omega^2$ . Thus, knowing that $g = GM/R^2$ , we have that the period is: $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{2R}{g}}$ Substituting the numbers given by the problem, one obtains the result.

time period of a long pendulum is given by $\frac { 2\pi }{ \sqrt { \left( \frac { 1 }{ R } +\frac { 1 }{ l } \right) g } }$ , use this formula and put the values of g and r effective on that planet !

Time period of a long pendulum is given by $\frac { 2\pi }{ \sqrt { \left( \frac { 1 }{ R } +\frac { 1 }{ l } \right) g } }$
g=GM/R^2 therefore g at the surface of planet will be g/2
l=2R
putting all values and solving we will get 6768